Calculus question...?

Question

how do I find all the open intervals on which the function f(x)= (x^2)/(x^2+4) is decreasing?

Answer 1

For any function, the intervals on which the function is decreasing correspond to the intervals on which its derivative is negative.

Using the quotient rule:

f'(x) = [(x² + 4)(2x) - (x²)(2x)]/(x² + 4)²
f'(x) = [2x³ + 8x - 2x³]/(x² + 4)²
f'(x) = 8x/(x² + 4)²

If you look at that function, the denominator is squared twice, so it will never be negative. So you only have to worry about the numerator, 8x. 8x is negative for x < 0 and positive for x > 0, so the function is decreasing on (-∞, 0) and increasing on (0, ∞).

Answer 2

For any function, the intervals on which the function is decreasing correspond to the intervals on which its derivative is negative.

Using the quotient rule:

f'(x) = [(x² + 4)(2x) - (x²)(2x)]/(x² + 4)²
f'(x) = [2x³ + 8x - 2x³]/(x² + 4)²
f'(x) = 8x/(x² + 4)²

If you look at that function, the denominator is squared twice, so it will never be negative. So you only have to worry about the numerator, 8x. 8x is negative for x < 0 and positive for x > 0, so the function is decreasing on (-∞, 0) and increasing on (0, ∞).

Answer 3

function f(x)= (x^2)/(x^2+4) = 1 - 4/(x^2+4)

f'(x) = 8x/(x^2+4)^2

On (-infinity, 0), f'(x) is negative. Therefore, f(x) is decreasing on (-infinity, 0).